Integration. 2. The Area Problem

Transcription

1 Integration Professor Richard Blecksmith Dept. of Mathematical Sciences Northern Illinois University richard/math2. Two Fundamental Problems of Calculus First Problem: Study how one variable changes with respect to another How does area change with respect to radius How does temperature change with respect to altitude How does y change with respect to x Second Problem: Find the area under a curve We use derivatives to solve Problem We use integrals to solve Problem 2 Given a curve y = f(x) 2. The Area Problem For now we will assume that f(x) is 0 How do we find the area under the curve and above the x-axis, between two vertical sides x = a and x = b? y = f(x) x = a Area x = b

2 2 3. A case we can do If the curve is linear, then we can use the formula for the area of a triangle A = 2 base height Example. y = f(x) = 2x between a = 0 and b = 5 base = 5 height = 0 A = = Graph of y = 2x (5, 0) y = 2x Area = 25 height=0 base = 5

3 3 5. A case we can t do (3, 9) (2, 4) y = f(x) = x 2 Area (, ) a = 0 b = 3 6. Rectangular Approximation y = x 2 3 rectangles

4 4 7. Rectangular Approximation y = x 2 6 rectangles 8. Rectangular Approximation y = x 2 2 rectangles 9. Sub-dividing [a, b] Set-up: divide the interval [a, b] into n equally spaced segments, whose division points are x 0, x,..., x n.

5 5 h a=x 0 x x 2 x 3 x 4 x n x n =b The width of each segment is h = b a n The division points x k are x 0 = a, x = a + h, x 2 = a + 2h, x 3 = a + 3h and, in general, x k = a + kh When a = 0, h 0. Special case: a = 0 0=x 0 x x 2 x 3 x 4 x n x n =b these forumlas become h = b n and x k = kh Note that x n = nh = n b n = b, as it should.. Riemann Method In the Riemann method, we use n rectangles to approximate the area under the curve y = f(x) Each rectangle has as its base the subinterval [x k, x k ] of width h The height of each rectangle is f(x k ) Since the area of a rectangle is base height, the k-th rectangle has area

6 6 h f(x k ) Our approximation, called a Riemann Sum is the sum of the areas of these n rectangles 2. Riemann Method Picture h 0=x 0 x x 2 x 3 x n x n=b 3. Using 3 subdivisions For the function f(x) = x 2, and interval [a, b] = [0, 3] suppose we use n = 3 rectangles We divide [0, 3] into 3 equally spaced subintervals, each of width h = 3/3 = with division points x k = kh = k 0=x 0 x = x 2 =2 x 3 = 3 Notice that the height of the kth rectangle with base [x k, x k ] is f(x k ) = x 2 k

7 7 k 2 3 x k 2 3 f(x k ) Rectangles Each rectangle has width h = and height f(x k ) = x 2 k The area of the 3 red rectangles is A(3) = h f(x ) + h f(x 2 ) + h f(x 3 ) = h x 2 + h x2 2 + h x2 3 = = = 4 5. Rectangular Approximation 9 y = x rectangles

8 Rectangles Each rectangle has width h =, subdivision point x 2 k = kh = k, and height 2 f(x k ) = x 2 k The area of the 6 bluish green rectangles is A(6) = h f(x ) + h f(x 2 ) + h f(x 3 ) + h f(x 4 ) + h f(x 5 ) + h f(x 6 ) = h x 2 + h x2 2 + h x2 3 + h x2 4 + h x2 5 + h x2 6 ( ) 2 ( 3 ) 2 ( 5 ) 2 = h + h () 2 + h + h (2) 2 + h + h (3) = = = = Using 6 Rectangles y = x 2 6 rectangles 8. Notation Time Out It is getting very tedious writing these sums

9 9 The sum with 2 rectangles will be even worse! To ease the writing, mathematicians invented the summation symbol It works like this: underneath we write the starting index above we write the ending index to the right of we write the form of the things we are adding 3 a k is the sum a + a 2 + a 3 k= 9. So What? It doesn t seem so bad to write these out! 50 a k is the sum k= a + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a 8 + a 9 + a 0 +a + a 2 + a 3 + a 4 + a 5 + a 6 +a 7 +a 8 +a 9 +a 20 +a 2 +a 22 +a 23 +a 24 +a 25 +a 26 +a 27 +a 28 +a 29 +a 30 +a 3 + a 32 + a 33 + a 34 + a 35 + a 36 + a 37 + a 38 + a 39 + a 40 +a 4 + a 42 + a 43 + a 44 + a 45 + a 46 + a 47 + a 48 + a 49 + a 50 Nobody wants to write all 50 terms. Though, in fairness we could have written this last sum as a + a 2 + a a 49 + a Summation Examples The sum of the first 0 integers can be written S = 0 k Evaluate this sum

10 0 S = 0 k = = Summation Examples What is S = 0 2k This is the sum of the first 0 even integers: S = = 0 55? Can you think of an easier way, using our previous calculation S = 0 k = Factor out the 2 from each term S = 0 2k = 2 0 k = 2 55 = Rectangular Approx of Area Using summation notation, we can write the sum of the areas of the 6 retangles used to approximate the area under the curve A(6) = h f(x ) + h f(x 2 ) + h f(x 3 ) + h f(x 4 ) + h f(x 5 ) + h f(x 6 ) as A(6) = 6 h f(x k ) To go from 6 to 2 rectangles is easy: A(2) = 2 h f(x k ) = h 2 f(x k ) by factoring out h.

11 23. Example y = x 2 Revisited y = f(x) = x 2 [a, b] = [0, 3] h = 3 n x k = hk A(n) = h n f(x k ) 24. Computing the exact area The following table gives the values of A(n) for various values of n: n A(n) It looks like the exact area under the curve y = x 2 between 0 and 3 is Calculating Areas The exact area A under the curve y = f(x) from a to b is defined to be the limit of the Riemann Sums A(n) as the number of rectangles n goes to infinity. That is, A = lim n A(n) = lim n h n f(x k ) k=

12 2 For many simple functions, such as f(x) = x n the limit that gives the area under the curve has been worked out. 26. Illuminating Example For example, suppose f(x) = x 2 and we wish to compute the area from a = 0 to b = x. It can be shown that ( A(n) = x3 + ) ( 2 + ) 6 n n so that x 3 ( A = lim A(n) = lim + ) ( 2 + ) = x3 x3 ( + 0) (2 + 0) = n n 6 n n 6 3 When x = 3, A = 33 3 = 9, a fact that we laboriously worked out earlier. 27. Area under y = x n Suppose f(x) = x n for any non-negative integer. We want to find the area A under the curve from a = 0 to b = x. Case: n = 0 Our function is just y = x 0 =. The region is one long rectangle, whose dimensions are x by. The area of this rectangle is A = x. Case: n = Our function is just y = x = x. 28. Area under y = x

13 (5, 0) 3 The region is an isosceles triangle whose dimensions are x by x. The area of this rectangle is A = 2 x2. y = x A = 2 x2 base = x height = x 29. Area under y = x 2 Case: n = 2 Our function is just y = x 2. We have already worked this out. The area of this region is A = 3 x3. a = 0 A = 3 x3 y = f(x) = x 2 b = x 30. Area under y = x n, n = 0,..., 6 For every possible value of the n, the area under the curve y = f(x) = x n from 0 to x has been calculated. These areas are given in the following chart:

14 4 f(x) x 0 = x x 2 x 3 x 4 x 5 x 6 A x 2 x2 3 x3 4 x4 5 x5 6 x6 7 x7 Differentiate the second column and compare with the first column. It looks like 3. Big Surprise! the derivative of the area is just the function f(x) That is, if A(x) = the area under the curve y = f(x) from a to x, then A (x) = f(x) So in order to solve the area problem, we must find a function whose derivative is f(x) Such functions are called anti-derivatives 32. Anti-derivatives We say that F (x) is an anti-derivative of f(x) if F (x) = f(x). Notice that the letters are case-sensitive. Essentially, you worked out the derivative of a function F (x), got the answer f(x). Can you figure out what was the original function just from knowing its derivative? The answer is sometimes Suppose you know

15 5 F (x) = 3x 2 4x + What s F (x)? F (x) = x 3 7x 2 + x works But so does F (x) = x 3 7x 2 + x The trouble with constants Since the derivative of any constant is zero, you cannot know the value of that constant just from the derivative. The following fact shows that, except for an undetermined constant, antiderivatives are essentially unique: If two functions f(x) and g(x) have the same derivative, then f(x) = g(x)+ C, for some constant C. When asked to find the antiderivative of F (x) = 3x 2 4x + the most general answer is to write F (x) = x 3 7x 2 + x + C where C is a constant. Mathematicians write the symbol 34. Integral Notation f(x) dx to mean the anti-derivative of f(x) The symbol is called the integral sign and the anti-derivative of f(x) is also called the integral of f(x). Thus,

16 6 3x 2 4x + dx = x 3 7x 2 + x + C For any value of n except, 35. The Power Rule x n dx = n + xn+ + C You can easily check this formula by differentiation: d dx n + xn+ + C = (n + )x(n+) n + = x n More good news: this formula works for rational numbers, negative numbers, and real numbers, as well as integers. x 0 dx = x + C x dx = 3 x dx = 36. Examples x 3 dx = 2 x 2 + C = 2 x 2 + C x /2 dx = 3/2 x3/2 + C = 2 3 x3/2 + C 37. Two Rules We can integrate sums such as x 3 + x by integrating each term separately: x 3 + x dx = x 3 dx + x dx = 4 x4 + 2 x2 + C

17 7 Why don t we write C + C? The rule that justifies this is f(x) + g(x) dx = f(x) dx + g(x) dx There is also a rule for pulling out constants : a f(x) dx = a f(x) dx where a is a constant 38. More Practice 5x dx = 5 3 x3 + 2x + C 6 4 x dx = 6 4x 2 dx = 6x 4 2 x + C = 6x + 4 x + C 6 3 x dx = 6x /3 dx = 6 4/3 x4/3 + C = 9 2 x4/3 + C 0x + 2 x + dx = = 5x 2 + 4x /2 + x + C 0x + 2x /2 + dx = 0 2 x2 + 2 /2 x/2 + x + C x 2 x Integrate dx x We first simply the the fraction x 2 x x(x ) dx = dx = x x 39. A quotient x dx = 2 x2 x + C Question: Could we have integrated this quotient by some kind of quotient rule for integrals?

18 8 40. The Bad News about Integration There is no product rule, quotient rule, integrals. Integrate (x 3 ) 2 dx or chain rule for There is no chain rule for integrals, so we have to square out the binomial: (x 3 ) 2 dx = x 6 2x 3 + dx = 7 x7 2 4 x4 + x + C